Question

let $n$ be an odd natural number such that the variance of $1,2,3,4,\dots ,n$ is $14.$ Then $n$ is equal to

Answer 1

Given: variance $=14$
$\Rightarrow \frac{\sum _{i=1}^n{x}_i^2}{n}-(\overline{x}{)}^2=14$
$\Rightarrow \frac{1^2+2^2+\dots +n^2}{n}-{\left(\frac{1+2+3+\dots +n}{n}\right)}^2=14$
$\Rightarrow \frac{\frac{n(n+1)(2n+1)}{6}}{n}-{\left(\frac{\left(\frac{n(n+1)}{2}\right)}{n}\right)}^2=14$
$\Rightarrow \frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4}=14$
$\Rightarrow \frac{(n+1)}{2}(\frac{(2n+1)}{3}-\frac{(n+1)}{2})=14$
$\Rightarrow (n+1)(4n+2-3n-3)=168$
$\Rightarrow n^2-1=168$
$\therefore n=13$